∫x∧5(sinx)∧2dx

2个回答

  • 求不定积分∫x⁵sin²xdx

    原式=(1/2)∫x⁵(1-cos2x)dx=(1/2)[∫x⁵dx-∫x⁵cos2xdx]=(1/2)[(1/6)x⁶-(1/2)∫x⁵dsin(2x)]

    =(1/12)x⁶-(1/4)[x⁵sin2x-5∫x⁴sin2xdx]=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)∫x⁴d(cos2x)

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)[x⁴cos2x-4∫x³cos2xdx]

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)∫x³d(sin2x)

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)[x³sin2x-3∫x²sin2xdx]

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)∫x²d(cos2x)

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)[x²cos2x-(1/2)∫xcos2xdx]

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)x²cos2x-(15/32)∫xd(sin2x)

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)x²cos2x-(15/32)[xsin2x-(1/2)∫sin2xd(2x)]

    =(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)x²cos2x-(15/32)xsin2x-(15/64)cos2x+C